MECHANICAL VIBRATIONS

A Lecturebook

Charles M. KrousgrillJe↵rey F. Rhoads

c� 2018 Charles M. Krousgrill and Je↵rey F. Rhoads

About the Authors:

Charles M. Krousgrill: Charles M. Krousgrill is a Professor in the School of Mechanical Engineer-ing at Purdue University. He received his M.S. in mechanical engineering from Purdue Universityin 1975, and his M.S. and Ph.D. in applied mechanics from the California Institute of Technologyin 1976 and 1980, respectively. During his more than 30 years at Purdue University, few, if any,individuals have had a greater impact on the university’s undergraduate students and the institu-tion’s commitment to engineering education. Dr. Krousgrill’s e↵orts in this regard have garnerednumerous awards and international accolades. To date, these include the Purdue University Schoolof Mechanical Engineering’s Harry L. Solberg Best Teacher Award (eight times), the Purdue Uni-versity College of Engineering’s Potter Best Teacher Award (four times), the Purdue UniversityMurphy Best Teacher Award, the 2010 Purdue University Helping Students Learn Award, the 2010Purdue Alumni Association Special Boilermaker Award and the American Society of EngineeringEducation’s 2011 Archie Higdon Distinguished Educator Award – the de facto lifetime achieve-ment award for educational accomplishments in the field of mechanics. In January 2018, it wasannounced that Dr. Krousgrill had been awarded the named 150th Anniversary Professorship byPurdue University. Outside of the classroom, Dr. Krousgrill conducts research in the general areasof dynamics and mechanical vibration.

Je↵rey F. Rhoads: Je↵rey F. Rhoads is a Professor in the School of Mechanical Engineering atPurdue University and is a�liated with both the Birck Nanotechnology Center and Ray W. HerrickLaboratories at the same institution. He also serves as the Director of Practice for MEERCatPurdue: The Mechanical Engineering Education Research Center at Purdue University and theAssociate Director of PERC: The Purdue Energetics Research Center. Dr. Rhoads received his B.S.,M.S., and Ph.D. degrees, each in mechanical engineering, from Michigan State University in 2002,2004, and 2007, respectively. Dr. Rhoads’ current research interests include the predictive design,analysis, and implementation of resonant micro/nanoelectromechanical systems (MEMS/NEMS)for use in chemical and biological sensing, electromechanical signal processing, and computing;the dynamics of parametrically-excited systems and coupled oscillators; the thermomechanics ofenergetic materials (including explosives, pyrotechnics, and propellants); additive manufacturing;and mechanics education. Dr. Rhoads is a Member of the American Society for EngineeringEducation (ASEE) and a Fellow of the American Society of Mechanical Engineers (ASME), wherehe serves on the Design Engineering Division’s Technical Committee on Vibration and Sound. Dr.Rhoads is a recipient of numerous research and teaching awards, including the National ScienceFoundation’s Faculty Early Career Development (CAREER) Award; the Purdue University Schoolof Mechanical Engineering’s Harry L. Solberg Best Teacher Award (twice), the Robert W. FoxOutstanding Instructor Award, the and B.F.S. Schaefer Outstanding Young Faculty Scholar Award;the ASEE Mechanics Division’s Ferdinand P. Beer and E. Russell Johnston, Jr. Outstanding NewMechanics Educator Award; and the ASME C. D. Mote Jr., Early Career Award. In 2014, Dr.Rhoads was included in ASEE Prism Magazine’s 20 Under 40.

ChapterI

Chapter I

Equations of motion for discretesystems

ChapterI

Introduction

In this chapter we will discuss several di↵erent approaches for obtaining the equations of motion(EOM’s) of systems for which the mass, damping and sti↵ness components appear in discretecomponents. These EOM’s will be ordinary di↵erential equations which describe the motion of thesystem. For the most part, we will be interested in describing small-amplitude motion. Becauseof this, we will be able to deal with a linearized form of these di↵erential equations, where thelinearization is performed about equilibrium states.

We will be considering a number of approaches in deriving the EOM’s:

• Newton-Euler formulation – This is a vector-based approach beginning with forces and ac-celerations written in vector form. In doing so, we will need to deal with all of the forcesacting on the system. In particular, we will include forces of reaction and forces of constraintalong with the applied forces. It is often the case that we will not care about quantifyingthe reaction and constraint forces, and consequently, we will eliminate these forces from theEOM’s before attempting to solve the EOM’s. This elimination can be a tedious task.

• Power equation – The power equation, as we will see, is based on a work/energy (scalar)description of the motion. An important consequence of this is that many of the forces ofreaction and constraint will naturally not appear in the energy equation. However, since wewill start with a single work/energy equation for the system of interest, we will obtain onlya single EOM regardless of the number of coordinates needed to completely describe the mo-tion. Hence, the power equation will be useful to us only for systems having a single degreeof freedom (DOF), that is, systems for which only one coordinate is needed to describe themotion.

• Lagrange’s equations – The Lagrangian formulation is a means by which we will be able toseparate out from the power equation the correct number of EOM’s needed to describe themotion of the system. That is, for a system having N DOFs, the Lagrangian formulation willproduce N EOM’s.

• Linearized equations of motion – For small amplitude oscillations about an equilibrium state,we will be able to replace nonlinear terms appearing in the EOM’s obtained from the La-grangian approach by their linearized approximations. This will produce a set of linearordinary di↵erential equations that we will use later on for determining the response of thesystem. We will be able to obtain these linear EOM’s directly from the kinetic energy, poten-tial energy and Rayleigh dissipation functions without directly using Lagrange’s equations.This formulation will allow us to observe symmetry properties of the mass and sti↵ness ma-trices. These symmetry properties will prove useful to use later on in the course in when weuse orthogonality properties of the modal vectors of the problem.

• Flexibility matrix and influence coe�cients – Later on in the course we will need to use the in-verse of the sti↵ness matrix (called the flexibility matrix) for some of the numerical eigenvalueextraction methods. Here we will discuss a direct approach for finding the flexibility matrixusing the so-called ’influence coe�cients. These influence coe�cients are generally easier tofind than the inverse of the sti↵ness and will provide some physical insight into the problem.

I.1 EOM’s: Newton-Euler equations

Background

Equations of Motion Using the Newton-Euler Equations

I.2 ME 563 – Fall 2002 Chapter I Notes

Background: For a single rigid body executing planar motion, we have the

following set of Newton-Euler equations:

F! = ma G

M A! = IA "

where G = center of mass of body A = EITHER center of mass OR fixed point of the rigid body

F! = resultant force vector acting on rigid body

M A! = resultant moment vector acting about point A IA = mass moment of inertia about point A

a G = acceleration vector of the center of mass G

! = angular acceleration vector of the rigid body

Remarks: a) ALWAYS draw appropriate free body diagrams (FBD’s) before you attempt

to develop the EOM’s for the system. b) Carefully consider the coordinates used (such as positive direction, whether

measured from fixed reference or moving reference, whether measured from equilibrium or unstretched positions) before drawing FBD’s.

c) You need to include ALL forces acting on the body when writing down the EOM’s (including forces of reaction). Later on when we use the power equation and Lagrange’s equations we will be able to ignore forces that do no virtual work when deriving EOM’s.

d) You should have exactly N EOM’s for a system having N degrees of freedom (we will discuss the concept of number of degrees of freedom in a later lecture). Typically you will have more than N EOM’s; however, you will be able to eliminate forces of constraint (reaction forces) and enforce kinematic constraints to reduce this number of EOM’s to N.

e) Never attempt to write down the EOM’s by inspection.

A

G

For a single rigid body executing planar motion, we have the following set of Newton-Euler equa-tions:

X~F = m~aG

X~MA = IA~↵

where

G = center of mass of the body

A = EITHER the center of mass OR a fixed point on the rigid bodyX

~F = the resultant force acting on the bodyX

~MA = the resultant moment about point A acting on the body

IA = the mass moment of inertia of the body about point A

~aG = the acceleration vector for the center of mass G

~↵ = the angular acceleration vector for the body

Remarks:

• ALWAYS draw appropriate free body diagrams (FBD’s) before you attempt to develop theEOM’s for the system.

• Carefully consider the coordinates used (such as sign conventions, whether measured fromfixed reference or moving reference, whether measured from equilibrium or unstretched posi-tions) before drawing FBD’s.

• You need to include ALL of the forces acting on the body when writing down the EOM’s(including forces of reaction). Later on when we use the power equation and Lagrange’sequations we will be able to ignore forces that do no virtual work when deriving EOM’s.

ChapterI

• In the end, you should have exactly N EOM’s for a system having N degrees of freedom (wewill discuss the concept of number of degrees of freedom in a later lecture). Typically youwill have more than N Newton/Euler equations at the start; however, you will be able toeliminate forces of constraint (reaction forces) and enforce kinematic constraints to reducethis number of EOM’s to N .

• Never attempt to write down the EOM’s by inspection.

Example I.1.1

A homogeneous cylinder having a mass of M and radius r rolls without slipping on an inclinedsurface. A spring having a sti↵ness of k is attached between the center G of the cylinder and ground.

• Determine the equation of motion (EOM) corresponding to the coordinate x, where x de-scribes the position on the incline for point G with x being measured from the positionof G when the spring is unstretched.

• Determine the EOM corresponding to the coordinate z, where z describes the position onthe incline for point G with z being measured from the equilibrium position of G.

x z

no slip no slip

θθ

unstretched position

equilibrium position

m m

r r

figure&a)& figure&b)&

k k

G G

ChapterI

Example I.1.2

Two particles, 1 and 2, (each of mass m) are interconnected by a set of three springs (each of sti↵-ness k) and three dashpots (each having a damping constant of c) and slide on a smooth horizontalsurface. A force f acts on particle 2.

Find the equations of motion (EOM’s) for this problem corresponding to three di↵erent sets ofcoordinates:

• Using x1 and x2 with both x1 and x2 being measured positive to the right from fixed referencepoints with the springs being unstretched when x1 = x2 = 0.

• Using y1 and y2 with y1 being measured positive to the right from a fixed reference point, y2

being measured positive to the left from a fixed reference point and with the springs beingunstretched when y1 = y2 = 0.

• Using z1 and z2 with z1 being measured positive to the right from a fixed reference point,z2 representing the stretch/compression of the middle spring (z2 > 0 when the spring isstretched) and with the springs being unstretched when z1 = z2 = 0.

figure&a)& figure&b)&

x1

f 1 2

x2

figure&a)&

y1

f 1 2

y2

figure&b)&

z1

f 1 2

z2

figure&c)&

ChapterI

Example I.1.3

Particle A (having a mass of m) slides on a smooth horizontal surface with a thin homogeneous bar(having a length of L and mass M) being attached to A with a smooth pin, as shown. A spring anddashpot connect A to ground. The coordinate x describes the motion of A (x is measured positiveto the right and x = 0 when the spring is unstretched). The coordinate ✓ describes the orientationof the bar, with ✓ being positive for a counterclockwise rotation from the vertical.

Find the EOM’s of the system using the coordinates of x and ✓.

x

k

c

G

A

θ

ChapterI

I.2 EOM’s: Power equation

Objectives

It is desired to be able to obtain the di↵erential equation of motion for a single degree-of-freedomsystem without having to account for “workless” forces acting on the system.

Background

The work-energy equation for a system of n rigid bodies is given by:

T + U = T0 + U0 + W (nc)

where

Tj = kinetic energy of the jth body =1

2mvAj

2 +1

2IAj!

2

T = kinetic energy of the system =nX

i=j

Tj

U = potential energy due to conservative forces

W (nc) = work done by nonconservative forces =X

j

Z 2

1

~F (j) · d~⇢j

Aj = EITHER a fixed point OR the center of mass of the jth body

vAj = speed of point Aj

! = angular speed of the body

T0, U0 = initial values for the kinetic and potential energies, respectively

Aj

j

ρj

O (fixedpoint)

pa th of point j

dρj (ta ngent to pa th)

ChapterI

Results

The “power equation” results from taking a time derivative of the work-energy equation:

Power =dW (nc)

dt=

dT

dt+

dU

dt

Remarks

• As we will see in the following examples, the power equation will produce a second-order dif-ferential EOM for a system for which only a single coordinate is needed to describe its motion(i.e., systems having a single degree of freedom). The power equation has the advantage overthe Newton-Euler approach in that one can ignore forces that do no nonconservative work onthe system.

• A particle is a rigid body for which either the dimensions are small (IG ⇡ 0) or is under puretranslation (! = 0). Therefore, the kinetic energy for a particle is given by T = 1

2mvG2.

• Note that the initial values for the kinetic and potential energies (T0 and U0, respectively)drop out when di↵erentiating the work-energy with respect to time.

Example I.2.1

A homogeneous thin bar of mass M and length L is pinned to ground at point O. Using the powerequation, find the equation of motion (EOM) for the bar corresponding to the coordinate of ✓ where✓ is measured counterclockwise from the vertical.

G

θ

g O

ChapterI

Example I.2.2

Use the power equation to find the EOM for the system shown corresponding to the coordinate x,where the coordinate x is defined such that x = 0 when the spring is unstretched.

x

m k

c f (t)

ChapterI

Example I.2.3

Use the power equation to find the EOM of the system shown corresponding to the coordinate �.The mass of the particle is M , and the mass of the homogeneous cylinder (outer radius r) is m.The bar connecting the particle and the center of the cylinder has a mass that is negligible. Thespring is unstretched when � = 0.

k

φ

f (t)

no slip smooth r M

m negligible ma ss O

C

ChapterI

I.3 EOM’s: Generalized coordinates, generalized forces and gen-eralized mass coe�cients

Objectives

Here we will rewrite the power equation from the last section in a way that the set of chosencoordinates and their time derivatives are explicitly apparent. In this form, sets of “generalizedforces” and ”generalized mass coe�cients” are introduced. The results of this section will lead usdirectly in the Lagrangian formulation of the EOM’s found in the next section.

Background

Ak

rk

O (fixedpoint)

θk

• For a system of n planar rigid bodies, let Ak be either the center of mass or a fixed point onthe kth body. Assume that the position vectors, ~rk, and angular rotations, ✓k, are writtenin terms of a set of N generalized coordinates qi (i = 1, 2, ..., N). It is assumed that thesegeneralized coordinates completely describe the configuration of the system for all time. It isalso assumed that ~rk and ✓k are not explicit functions of time. From this, we can write fork = 1, 2, ..., n:

~rk = ~rk (q1, q2, ..., qN )

✓k = ✓k (q1, q2, ..., qN )

• For a function b = b (q1, q2, ..., qN ), we have the chain rule of di↵erentials:

db =@b

@q1dq1 +

@b

@q2dq2 + ... +

@b

@qNdqN =

NX

i=1

@b

@qidqi

• For a function b = b (q1, q2, ..., qN ), we have the chain rule of di↵erentiation:

db

dt= b =

@b

@q1q1 +

@b

@q2q2 + ... +

@b

@qNqN =

NX

i=1

@b

@qiqi

ChapterI

Results

T =1

2

NX

i=1

NX

k=1

mikqiqk

dT =NX

i=1

d

dt

✓@T

@qi

◆� @T

@qi

�dqi

dU =NX

i=1

@U

@qidqi

dW (nc) =NX

i=1

Qidqi

where Qi = “generalized force” corresponding the generalized coordinate qi and mik = “generalizedmass coe�cient”, each corresponding to the generalized coordinates qi and qk and given by:

Qi =MX

j=1

~Fj · @~⇢j@qi

mik =nX

j=1

mj

@~rj@qi

· @~rj@qk

+ Ij@✓j@qi

· @✓j@qk

�

where M is the number of applied forces on the system.

Derivation

Recall the following set of Newton-Euler equations for planar motion of a rigid body:

~Fj = mj~aGj = mj~rj

Mj = Ij↵j = Ij ✓j

where

~Fj = the resultant force acting on the jth body

Mj = the resultant moment about point Aj acting on the jth body

Gj = center of mass of the jth body

Aj = EITHER the center of mass OR a fixed point on the jth rigid body

Say we take the dot product of the first equation with ~rj , multiply the second equation by d✓j , sumthose results over all n bodies and add the results, producing:

nX

j=1

hmj~rj · d~rj + Ij ✓jd✓j

i=

nX

j=1

h~Fj · d~rj + Mjd✓j

i

We recognize the right hand side of the above equation as being the total di↵erential work doneon the system. Alternately, we can write this right hand side of the equation as the di↵erencebetween the di↵erential work done by nonconservative forces and the di↵erential of the potentialof conservative forces:

nX

j=1

h~Fj · d~rj + Mjd✓j

i= dW (nc) � dU

Therefore, the left hand side of this equation, according to the work-energy equation,

dT + dU = dW (nc)

must be the di↵erential kinetic energy for the system:

dT =nX

j=1

hmj~rj · d~rj + Ij ✓jd✓j

i

In this section of the notes, we will work to simplify the expressions for the three terms of dT , dUand dW (nc). Consider the following steps:

• From the chain rule of di↵erentials:

d~rj =NX

i=1

@~rj@qi

dqi

d✓j =NX

i=1

@✓j@qi

dqi

ChapterI

Substituting these expressions into the expression for dT gives:

dT =nX

j=1

"mj~rj ·

NX

i=1

@~rj@qi

dqi

!+ Ij ✓j

NX

i=1

@✓j@qi

dqi

!#

=NX

i=1

2

4nX

j=1

✓mj~rj · @~rj

@qi+ Ij ✓j

@✓j@qi

◆3

5 dqi

Note that through the product rule of di↵erentiation:

~rj · @~rj@qi

=d

dt

✓~rj · @~rj

@qi

◆� ~rj · d

dt

✓@~rj@qi

◆

=d

dt

✓~rj · @~rj

@qi

◆� ~rj · @~rj

@qi

It can be shown that (see Appendix I):

@~rj@qi

=@~rj@qi

Substituting into the above gives:

~rj · @~rj@qi

=d

dt

~rj · @~rj

@qi

!� ~rj · @~rj

@qi

=1

2

d

dt

@

@qi

⇣~rj · ~rj

⌘�� 1

2

@

@qi

⇣~rj · ~rj

⌘

In a similar way, it can be shown that:

✓j@✓j@qi

=1

2

d

dt

"@✓2

j

@qi

#� 1

2

@✓2j

@qi

Substituting these two results into the above equation for dT gives:

dT =NX

i=1

d

dt

2

4 @

@qi

nX

j=1

✓1

2mj~rj · ~rj +

1

2Ij ✓

2j

◆3

5 dqi �NX

i=1

2

4 @

@qi

nX

j=1

✓1

2mj~rj · ~rj +

1

2Ij ✓

2j

◆3

5 dqi

=NX

i=1

d

dt

✓@T

@qi

◆� @T

@qi

�dqi

where:

T =nX

j=1

✓1

2mj~rj · ~rj +

1

2Ij ✓j

2◆

• Using the chain rule of di↵erentiation:

T =1

2

nX

j=1

mj~rj · ~rj +1

2

nX

j=1

Ij ✓j2

=1

2

nX

j=1

mj

NX

i=1

@~rj@qi

qi

!·

NX

k=1

@~rj@qk

qk

!+

1

2

nX

j=1

Ij

NX

i=1

@✓j@qi

qi

! NX

k=1

@✓j@qk

qk

!

=1

2

NX

i=1

NX

k=1

0

@nX

j=1

mj

@~rj@qi

· @~rj@qk

1

A qiqk +1

2

NX

i=1

NX

k=1

0

@nX

j=1

Ij@✓j@qi

@✓j@qk

1

A qiqk

=1

2

NX

i=1

NX

k=1

mikqiqk

where

mik =nX

j=1

mj

@~rj@qi

· @~rj@qk

+ Ij@✓j@qi

@✓j@qk

�

are the “generalized mass coe�cients”. Two things to note from the above. First, the kineticenergy expression T is a function of both the N generalized coordinates qj (through the mik

coe�cients) and their time derivatives qj . Secondly, the coe�cients mik are “symmetric”;that is, mik = mki.

• The potential energy is a function of only the spatial configuration of the system. Therefore,U can be expressed completely in terms of the generalized coordinates qi and will not involvetheir time derivatives qi: U = U (q1, q2, ..., qN ). Using the chain rule of di↵erentials, we cantherefore write:

dU =NX

i=1

@U

@qidqi

• Suppose that we have M nonconservative forces ~Fj (j = 1, 2, ..., M) acting at the followingM locations within the system: ~⇢j = ~⇢j (q1, q2, ..., qN ). Using the chain rule of di↵erentialsgives us:

d~⇢j =NX

i=1

@~⇢j@qi

dqi

ChapterI

Using this in the di↵erential work term produces:

dW (nc) =MX

j=1

~Fj · d~⇢j

=MX

j=1

~Fj ·

NX

i=1

@~⇢j@qi

dqi

!

=NX

i=1

0

@MX

j=1

~Fj · @~⇢j@qi

1

A dqi

=NX

i=1

Qidqi

where

Qi =MX

j=1

~Fj · @~⇢j@qi

= “generalized force” corresponding to the generalized coordinate qi

• In summary, the di↵erential form of the power equation:

dT + dU = dW (nc)

can be written in the following explicit form:

NX

i=1

d

dt

✓@T

@qi

◆� @T

@qi+

@U

@qi� Qi

�dqi = 0

Example I.3.1

Forces f1 , f2 and f3 act on the three-particle system shown. Generalized coordinates x1, x2 andx3 are used to describe the motion of these particles, where x2 and x3 are relative coordinates.

• Find the generalized forces corresponding to the generalized coordinates x1, x2 and x3 for theforces f1, f2 and f3 .

• Find the generalized mass coe�cients corresponding to the generalized coordinates x1 , x2

and x3 .

x1

f2 1 2

x2

f3 3

m m m

x3

f1

x

y

ChapterI

I.4 EOM’s: Lagrange’s equations

Objectives

Our goal here is to develop a systematic method for deriving the equations of motion for a systemhaving N degrees of freedom. We will start with the explicit form of the power equation derivedin the last section. From the power equation we will obtain a set of N EOM’s for the system.

Background

In the last section of the notes, we saw that the power equation for a system described by Ngeneralized coordinates qi (i = 1, 2, ..., N) can be written as:

NX

i=1

d

dt

✓@T

@qi

◆� @T

@qi+

@U

@qi� Qi

�dqi = 0

where T and U are the kinetic and potential energies for the system, and Qi is the generalized forcecorresponding to the coordinate qi:

Qi =MX

j=1

~Fj · @~⇢j@qi

ChapterI

Derivation and results

It is assumed that the generalized coordinates qi (i = 1, 2, ..., N) used above completely describe theconfiguration of the system at all instances in time. However, we have said nothing at this pointas to whether the coordinates chosen are independent. That is, some of the coordinates could berelated by constraints that have not yet been enforced, and as a result, the coordinates are notindependent.

Consider an example of the simple pendulum shown below. The kinetic and potential energies for

!

L

x

y m

!

P

this system in terms of the (x, y) coordinates for the particle P are given by:

T =1

2m�x2 + y2

�=

1

2m�q21 + q2

2

�

U = �mgy = �mgq2

where q1 = x and q2 = y. Note that the chosen generalized coordinates are related by the constraintq21 + q2

2 = L2, which by di↵erentiating with respect to time produces: q21 = q2

2 q22/�L2 � q2

2

�. Al-

though q1 and q2 completely describe the configuration of the system, they are NOT independent.If we enforce the above constraint, T and U can be written in terms of q2 alone as:

T =1

2m

q22

L2 � q22

+ 1

�q22

U = �mgq2

This single degree-of-freedom system is now described in terms of a single coordinate.

An alternate (and better) choice of coordinates would be to describe the system in terms of theangle ✓. From the figure we see that we have the following constraints: x = Lsin✓ and y = Lcos✓.Substituting these constraints into the original T and U gives:

T =1

2mhL2✓2cos2✓ + L2✓2cos2✓

i=

1

2mL2✓2 =

1

2mL2q2

3

U = �mgLcos✓ = �mgLcosq3

where q3 = ✓. Again, we have used a single coordinate in describing the motion of a single degree-of-freedom system.

At this point, let us assume that all of the generalized coordinates are independent: any givencoordinate cannot be expressed in terms of the remaining N -1 coordinates. With this being thecase, N now represents the total number of degrees of freedom in the system. Since the coordinatesare independent, the general form of our power equation:

NX

i=1

d

dt

✓@T

@qi

◆� @T

@qi+

@U

@qi� Qi

�dqi = 0

becomes N independent EOM’s of the form:

d

dt

✓@T

@qi

◆� @T

@qi+

@U

@qi= Qi

for i = 1, 2, ..., N . The above are known as the set of Lagrange’s equations for an N degree-of-freedom system.

Recall that the potential energy term U includes the work done by conservative forces (such asgravitational and spring forces). The generalized force terms Qi includes the work done by all otherforces. The contribution of damping terms naturally appears within the generalized force terms.It is possible to make the contribution of damping forces more explicit in Lagrange’s equationsthrough a so-called Rayleigh dissipation function R. The Rayleigh dissipation function for a singlelinear dashpot can be written as:

Rj =1

2cj�

2j

where cj is the damping coe�cient and �j is the relative speed across the two ends of the dashpot.If the system has r dashpots, then the total Rayleigh dissipation function is:

R =rX

j=1

Rj =1

2

rX

j=1

cj�2j

Including damping produces the following modified form of Lagrange’s equations:

d

dt

✓@T

@qi

◆� @T

@qi+

@R

@qi+

@U

@qi= Qi

for i = 1, 2, ..., N . Here, the generalized forces Qj include all nonconservative forces except thosefrom the damping terms included in R.

ChapterI

Steps in using Lagrange’s equations

1. Number of degrees of freedom (DOFs) Determine the number of degrees of freedom(DOF’s) in the problem. To do so, carefully consider the least number of generalized coor-dinates N that are needed to completely describe the configuration of the system at any time.

2. Generalized coordinates. Choose your set of generalized coordinates qj(t) for j = 1, 2, ..., N .Be sure that you have chosen an independent set of coordinates. (Ask the question: Can youchange each coordinate individually while holding the other coordinates fixed and not violateany motion constraints on the system?) If your coordinates are not independent, there areconstraints that must exist among your coordinates. Enforce the constraints at this pointbefore proceeding. Also, reconsider your decision on the number of DOF’s in 1. above beforecontinuing: What is the correct number of DOFs?

3. Free body diagrams. Draw free body diagrams (FBD’s) for all bodies in your system.These FBD’s will be necessary later on when you derive the generalized forces acting on thesystem.

4. Kinetic energy expression, T.• Write down the velocity vector corresponding to point Aj (either center of mass or fixed

point) for each body (j = 1, 2, ..., n):

~vj =d~rjdt

If Aj is a fixed point, then, of course, ~vj = 0.• Write down the angular velocity corresponding to each body (j = 1, 2, ..., n):

!j =d✓jdt

• Form the kinetic energy expression for each body:

Tj =1

2mj~vj · ~vj +

1

2Ij!

2j

(Note that the first term in the above expression is a dot product of the velocity vectorwith itself. Take care to write down the vector expression for velocity before writingdown the kinetic energy expression. Also, Ij is the mass moment of inertia of the jthbody for point the choice of point Aj . If the body is to be treated as a particle, thenIj = 0.)

• Form the total kinetic energy for the system by summing up the kinetic energy expres-sions for the n rigid bodies:

T =nX

j=1

Tj

• Expand the expression for T to explicitly show the appearance of the qj terms:

T =1

2

NX

i=1

NX

k=1

mikqiqk

From this, identify the mass matrix elements mik. This step will help to simplify thedi↵erentiation process later on.

5. Potential energy expression, U.• Write down the potential energy for each spring in the system:

(Usp)j =1

2kj�

2j

• Write down the gravitational potential energy for each body in the system:

(Ugr)j = mjghj

where hj is the elevation of the center of mass of the body above the datum line thatyou have chosen. Please note that if hj > 0, the center of mass is above the datum line,whereas hj < 0 corresponds to the center of mass below the datum line.

• Form the total potential energy for the system:

U =X

(Usp)j +nX

j=1

(Ugr)j

6. Rayleigh dissipation expression, R.• Write down the Rayleigh dissipation for each dashpot in the system:

Rj =1

2cj�j

2

Form the total Rayleigh dissipation function for the system:

R =rX

j=1

Rj

7. Generalized forces, Qi.• Reconsider your FBD’s of the system. Determine which forces that do virtual work on

the system. A reminder of some forces that you will NOT include:– Forces due to springs, gravitation attraction and dashpots (since you have already

included these in U and R).– Contact forces at smooth interfaces. These will do no work on the system.– Frictional forces required for rolling without slipping. These do no work on the

system.

ChapterI

– Constraint forces due to rigid connections between bodies. These do work on theindividual bodies but NOT on the entire system.

• Write down position vectors for the point of application of forces that do virtual workon the system:

~⇢j = ~⇢j(q1, q2, ..., qN )

• Find the di↵erential change in the position vectors for the point of application of theforces:

d~⇢j =NX

i=1

@~⇢j@qi

dqi

• Find the di↵erential work for each force:

dWj = ~Fj · d~⇢j = ~Fj ·NX

i=1

@~⇢j@qi

dqi

• Find the di↵erential work for entire system, expand and identify the generalized forcesQi (i = 1, 2, , N) as the coe�cients of the individual dqi terms in the following expressionfor dW :

dW =MX

j=1

Wj =NX

i=1

Qidqi

8. Equations of motion.• Form the partial derivatives of T with respect to qi and qi (i = 1, 2, , N) . Recall that in

forming the partial derivatives with respect to variables qi and qi, only the EXPLICITappearance of the variables are a↵ected by the partial di↵erentiation.

• Form the total time derivative for the terms:

d

dt

✓@T

@qi

◆

Here, when finding the total time derivative, both EXPLICIT and IMPLICIT (throughthe qi and qi terms) appearance of time t is a↵ected by the di↵erentiation.

• Form the partial derivatives of U and R with respect to qi and qi, respectively• Combine results to form the N Lagrange’s equations for the system (i = 1, 2, ..., N):

d

dt

✓@T

@qi

◆� @T

@qi+

@R

@qi+

@U

@qi= Qi

where N is the number of DOF’s in the system.

Concluding remarks on Lagrange’s equations: some advanced ideas

Recall that we made two important assumptions in deriving the preceding form of Lagrange’s equa-tions: i) the position vectors and rotation angles are not explicit functions of time (“scleronomic”systems), and ii) we have chosen an independent set of coordinates. The following discusses theconsequences of these assumptions and what needs to be done when the assumptions are not validfor a particular system of interest.

Rheonomic systems. “Rheonomic” systems are those for which the position vectors and rotationangles are explicit functions of time: ~rj = ~rj (q1, q2, ..., qN , t) and ✓j = ✓j (q1, q2, ..., qN , t). With theexplicit time dependence, the chain rule of di↵erentials gives us:

d~rj =NX

i=1

@~rj@qi

dqi +@~rj@t

dt

With this, we see that the di↵erential work for an applied force includes an additional term due tothis explicit time dependence:

dWj = ~Fj · d~⇢j = ~Fj ·"

NX

i=1

@~⇢j@qi

dqi +@~⇢j@t

dt

#

For rheonomic systems, Lagrange’s equations are typically developed in terms of “virtual” displace-ments �~rj , where:

�~rj =NX

i=1

@~rj@qi

�qi

Note that virtual displacements are di↵erential displacements where the explicit time dependenceis frozen while applying the chain rule. In fact, for scleronomic systems, virtual and di↵erentialdisplacements are exactly the same: �~rj = d~rj .

Using virtual displacements gives us “virtual work”, �Wj :

�Wj = ~Fj · �~⇢j = ~Fj ·"

NX

i=1

@~⇢j@qi

�qi

#=

NX

i=1

~Fj · @~⇢j

@qi

��qi =

NX

i=1

Qj�qi

and, as a result, the generalized forces Qi are the same as before. Therefore, the Lagrangian formu-lation developed in this section for scleronomic systems are still applicable for rheonomic systemsprovided that we use virtual work (rather than di↵erential work) in finding our generalized forces.

Example of a rheonomic systemOne of the most common instances of rheonomic systems are those on which “base excitation” isapplied. Consider the two-DOF system shown below where the base “B” of the system is given aprescribed motion of y(t) = y0sin!t. The motion of particles 1 and 2 is described by the relativecoordinates x1 and x2. With that, the position vectors for these two particles are given by:

~r1 = [y0sin!t + x1] i = ~r1 (x1, t)

ChapterI

~r2 = [y0sin!t + x1 + x2] i = ~r2 (x1, x2, t)

where we see here that due to the prescribed motion of base B, the position vectors for the twoparticles are explicit functions of time.

y(t)

f21

x1

f1 2

x2

Bi

j

Using the chain rule of di↵erentials:

d~r1 = [y0!cos!tdt + dx1] i

d~r2 = [y0!cos!tdt + dx1 + dx2] i

and the definition of a virtual displacement:

�~r1 = [�x1] i

�~r2 = [�x1 + �x2] i

we can write the work and virtual work done by the applied forces:

dU = ~f1 · d~r1 + ~f2 · d~r2 = (y0!cos!t) (f1 + f2) dt + (f1 + f2) dx1 + f2dx2

�U = ~f1 · �~r1 + ~f2 · �~r2 = (f1 + f2) �x1 + f2�x2

The di↵erence between the work and virtual work expressions above is that virtual work doesnot include the influence of the prescribed displacement of the base within the work expression,whereas it must be included in the work expression. This is how using virtual displacements inthe Lagrangian formulation simplifies the EOM’s and is the recommended formulation for derivingEOM’s for rheonomic systems.

Nonholonomic systems. Consider the following system made up of particles A and B. Particle

A B

knife − edge smooth

side view

m m A

B

θ

top view

vC

vB

j i

J

I

cart, C

A moves within a slot cut into cart C (of mass M), as shown, where C is moving with a prescribedspeed (vC) and direction (X). The slot is perpendicular to the prescribed direction of motion forC. Particle B is attached to A with a massless, rigid rod of length L. B rides along on the samehorizontal surface as C on a knife-edge support. This support allows for smooth sliding of B alongthe direction of the rod but does not allow slip perpendicular to the rod. Rotation about thecontact point of the knife-edge with the ground is possible. Enforcing these constraints on a rigidbody kinematics equation relating the velocities of A and B gives:

~vB = ~vA + ~! ⇥ ~rB/A

vB i = XAI + YAI +⇣✓k⌘

⇥⇣�Li

⌘

= XAI + YAI � L✓j

From the figure we see that:

I = cos✓i � sin✓j

J = sin✓i + cos✓j

Therefore:

vB i =⇣XAcos✓ + YAsin✓

⌘i +⇣�XAsin✓ + YAcos✓ � L✓

⌘j

Balancing coe�cients gives the following two constraints:

vB = XAcos✓ + YAsin✓ = vCcos✓ + YAsin✓

0 = �XAsin✓ + YAcos✓ � L✓ = �vCsin✓ + YAcos✓ � L✓

Using the first constraint equation above in the kinetic energy expression for the system gives:

T =1

2mv2

A +1

2mv2

B +1

2Mv2

C

=1

2m⇣v2C + Y 2

A

⌘+

1

2m⇣v2Ccos2✓ + Y 2

Asin2✓⌘

+1

2Mv2

C

ChapterI

From this we see that the kinetic energy is a function of both coordinates YA and ✓. The secondconstraint above shows that these coordinates are NOT independent; however, due to the natureof this constraint, we are not able to enforce it. (Do you see the problem? The constraint isnon-integrable; that is, we cannot relate ✓ to YA without actually solving the problem first.) TheLagrangian formulation developed here relies on independent coordinates, and therefore we cannotuse this form of Lagrange’s equations to develop the EOM for this single DOF system.

Systems having non-integrable constraints (such as the example above) are known as “nonholo-nomic”. Since we cannot enforce these constraints apriori, Lagrange’s equations cannot be appliedto such systems. There are a number of ways to modify Lagrange’s equations to handle nonholo-nomic systems. We will not pursue that here. You are encouraged to read advanced dynamicstextbooks if you are interested in learning more. Should we encounter such systems in this course,we will avoid this complication and develop the EOM’s through the Newton-Euler formulation.

Example I.4.1

Find the EOM for the simple pendulum using ✓ as the generalized coordinate.

L

θ

g O

x

y

P m

negligible ma ss

ChapterI

Example I.4.2

Particles 1 and 2 (having masses of m and M , respectively) move in a HORIZONTAL plane.Generalized coordinates x1 and x2 describe the position of the system where x2 is measured relativeto particle 1. Forces f and F act in the x-direction on particles 1 and 2, respectively. The springsare unstretched when x1 = x2 = 0. Find the EOM’s of the system using generalized coordinatesx1 and x2. Ignore gravity. Consider all of the surfaces to be smooth.

x1

x2

kθ

m

cK

C

M

Ff

1

2

x

y

ChapterI

Example I.4.3

Find the EOM’s of system shown below using x, � and ✓ as generalized coordinates. The springis unstretched when x = � = 0. All three bodies are to be considered to be homogeneous in theirmass distribution, with each body have a mass of m.

x

k

L

A

θ

O

r

no slip smooth

1 2

φ

g

3

x

y

ChapterI

Example I.4.4

If particle 0 is given a prescribed motion of y(t) = y0sin⌦t, find the EOM’s for the system. Notethat x1 and x2 are relative coordinates. The springs are unstretched when y = x1 = x2 = 0.

y(t)

f2 1

x1

f1 2

m m

x2

k k k

ChapterI

I.5 EOM’s: Linearization

Objectives

The EOM’s for a system are typically nonlinear, with the nonlinearities arising from geometric andmaterial e↵ects. In this course, we will study small oscillations of systems. For this, we will usea linearized set of free response EOM’s arising from Lagrange’s equations. In this section of thenotes, we will develop a systematic approach for this linearization process for free motion of thesystem.

Background

• The kinetic energy for discrete models is of the form:

T =1

2

NX

i=1

NX

k=1

mikqiqk

where the symmetric mass coe�cients mik are given by:

mik =nX

j=1

mj

@~rj@qi

· @~rj@qk

+ Ij@✓j@qi

@✓j@qk

�

• For free response, the system is at an equilibrium state ~q0 when (i = 1, 2, ..., N):✓

@U

@qi

◆

~q0

= 0

• The Taylor series expansion of a function g(~q) about the point ~q0 is given by:

g(~q) = g(~q0) +NX

i=1

@g

@qi

�

~q0

(qi � q0i) +1

2

NX

i=1

NX

k=1

@2g

@qiqk

�

~q0

(qi � q0i) (qk � q0k) + ...

Results

Small motion about an equilibrium state of a free system is governed by the following linearizedEOM’s:

[M ] ~z + [C] ~z + [K] ~z = ~0

where

~z(t) = ~q(t) � ~q0

Mik = (mik)~q0 = Mki

Cik =

✓@2R

@qi@qk

◆

~q0

= Cki

Kik =

✓@2U

@qi@qk

◆

~q0

= Kki

ChapterI

Derivation and remarks

Example I.5.1

In the system shown, the springs are unstretched when x1 = x2 = 0. Find the mass, damping andsti↵ness matrices for small motion about the equilibrium state corresponding to the generalizedcoordinates shown. Note that x2 is a relative coordinate.

Example I.5.1

In the system shown, the springs are unstretched when x1 = x2 = 0. Find the mass, damping andsti↵ness matrices for small motion about the equilibrium state corresponding to the generalizedcoordinates shown. Note that x2 is a relative coordinate.

x1

k

θ

f

x

y

x2

k

c c

m

2m

g2

1

L negligible mass

~r1 = x1i

~r2 = (x1 + x2)i

~r3 = (x1 + x2 + Lsin✓)i + (�Lcos✓)j

~v1 = x1i

~v2 = (x1 + x2)i

~v3 = (x1 + x2 + L✓cos✓)i + (L✓sin✓)j

3

~r1 = x1i

~r2 = (x1 + x2)i

~r3 = (x1 + x2 + Lsin✓)i + (�Lcos✓)j

~v1 = x1i

~v2 = (x1 + x2)i

~v3 = (x1 + x2 + L✓cos✓)i + (L✓sin✓)j

ChapterI

I.6 EOM’s: Influence coe�cients and the flexibility matrix

Objectives

Later on in our investigation of approximate methods for finding natural frequencies and modalvectors, we will find it necessary to obtain the inverse of the sti↵ness matrix [K]. This inverse,[K]�1, is known as the “flexibility matrix”. In this section, we will explore a method for obtainingthe flexibility matrix directly without the need for performing the inversion of an N ⇥ N matrix.

Background

• Consider a generalized force Qj = 1 (unit magnitude) corresponding to generalized coordinateqj . Let qi be the displacement of the ith generalized coordinate as a result of the applied forceQj . The “influence coe�cient” aij is defined as displacement of the generalized coordinate qi

qi q j

Qj

due to the application of a unit magnitude generalized force Qj .

• We have seen that the sti↵ness matrix for a conservative system obtained from the Lagrangianformulation is symmetric; that is, [K]T = [K]. Since the order of the transpose and inverseoperations on a non-singular matrix is reversible, it can be seen that [K]�1 is also symmetric:

⇣[K]�1

⌘T=⇣[K]T

⌘�1= [K]�1

Results

• The influence coe�cients aij form the elements of the flexibility matrix; that is,⇣[K]�1

⌘

ij

=

aij .

• For a linearly elastic system, the influence coe�cients aij defined above are symmetric: aji.In words, this says that a unit force applied at j will produce the same displacement at ias the displacement at j due to a unit force at i. This is known as “Maxwell’s reciprocitytheorem”.

ChapterI

• Method for determing the influence coe�cients:– Apply a unit generalized force Qj with Qk = 0 for k 6= j.

– Calculate/measure the change in the displacements of the generalized coordinates qi(i =1, 2, ...N). These displacements are the influence coe�cients aij(i = 1, 2, ...N).

– Use the symmetry of aij to simplify your calculations; that is, although a symmetricN ⇥ N matrix has N2 terms, it has only N(N + 1)/2 unique terms. Furthermore, aijmay be easier to compute than aji in some instances.

Derivation and remarks

Example I.6.1

Consider a homogeneous, clamped beam having a length of L and flexural rigidity of EI. Find theflexibility matrix for the beam using the transverse deflections at the three particles as generalizedcoordinates. Note that from mechanics of materials, we know that the deflection at x, u(x), due toa force F applied at location a � x is given by:

u(x) =F

6EI

h2a3 � 3a2 (a � x) + (a � x)3

i

Influence Coefficients and the Flexibility Matrix

(Sections 6.1-6.2)

I.26 ME 563 – Fall 2002 Chapter I Notes

Example 1 - flexibility matrix for cantilevered beam Consider a homogeneous, clamped beam having a length of 3L and flexural rigidity of EI. Find the flexibility matrix for the beam using the transverse deflections at the three particles as generalized coordinates.

Solution From mechanics of materials, we know that the deflection at x, u(x), due to a force F applied at location a is given by:

u(x) = F

6EI [2a3 - 3a2(a - x) + (a - x)3 ] ; x ≤ a

L/3 L/3 L/3

F

u(x) x

a

ChapterI

Example I.6.2

Find the flexibility matrix for the three-DOF system using the absolute coordinates shown. Thesprings are unstretched when x1 = x2 = x3 = 0.

x1 x2 x3

1 2 3k 2k k

smooth

ChapterI

I.7 EOM’s: Summary

Newton-Euler formulation• Vector approach dealing with each body individually.• Must deal with ALL forces acting on each body.• Often times the reduction of the equations to a number equal to the number of DOF’s can

be tedious.• Dealing with the signs and direction of spring forces can be tedious.

Power equations• Scalar approach dealing with the system as a whole.• Need not deal with forces which do no work on the system as a whole.• Dealing with the signs and direction of spring forces is more straight-forward.• Useful for single-DOF systems only.

Lagrange’s equations• Scalar approach dealing with the system as a whole.• Need not deal with forces which do no virtual work on the system as a whole.• Dealing with the signs and direction of spring forces is more straight-forward.• Valid for systems with any number of DOF’s provided that one uses independent generalized

coordinates.• Development of generalized forces is a straight-forward procedure when using expressions for

virtual work:– write down a vector expression for each force, ~Fj .– write down a vector expression for the di↵erential (or virtual) displacement for the point

of application of each force, d~⇢j– then from

dW =MX

j=1

~Fj · d~⇢j =MX

j=1

Qjdqj

we can identify the corresponding generalized forces Qj .

Linearized EOM’s• General form of the linearized EOM’s for free response: [M ] ~z + [C] ~z + [K] ~z = ~0• The mass and sti↵ness matrices are always symmetric (unlike those found from Newton-

Euler).• These EOM’s describe small motion about an equilibrium state: ~z(t) = ~q(t) � ~q0.